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Proof step 4 from proof at-put-keys-01 (previous step)

sks0 + sk ' ≠ [], ∀ sk, sk0, st, r, val, r0. sk = sk0 ∨ (st[r -- (sks0 + sk '), val][r0 -- sk0]) = st[r0 -- sk0] ⊦ sk = sk0, (st[st[r -- sk1] .refval -- (sks0 + sk '), val][r0 -- sk0]) = st[r0 -- sk0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ sk, sk0, st, r, val, r0. sk = sk0 ∨ (st[r -- (sks0 + sk '), val][r0 -- sk0]) = st[r0 -- sk0] ⊦ sk = sk0, (st[st[r -- sk1] .refval -- (sks0 + sk '), val][r0 -- sk0]) = st[r0 -- sk0]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sks + sk ' ≠ []])]]